Converting integral to cylindrical coordinates

Question

For systems that exhibit cylindrical symmetry, it is natural to perform integration in cylindrical coordinates $(r, \phi, z)$ The relations between cartesian coordinates and cylindrical coordinates are: $x= r\cos \phi$, $y= r\sin\phi$, $z=z$,

Then, convert the integral $\int_{-1}^1 \int_{0}^{\sqrt{1-y^2}}\int_{x^2+y^2}^{\sqrt{x^2+y^2}} xyz dz dx dy$

I've first calculated all the different partial derivative in order to evaluate the jacobian which gives $J= r$

The integration limits looks unusual. For example, for $x$, it goes from $0$ to $\sqrt{1-y^2}$

How do I visualise to see which quadrant is it in or the region of space am I looking at?

From here how do I change the integration limits from $xyz$ to $r, \phi, z$

Answer 1

Step 1: To know the region $E$.

By hypothesis of the problem we have the region $E$ given by $$E=\left\{(x,y,z):\underbrace{-1\leqslant y\leqslant 1, 0\leqslant x\leqslant \sqrt{1-y^{2}}}_{(*)}, x^{2}+y^{2}\leqslant z\leqslant \sqrt{x^{2}+y^{2}}\right\}$$

Step 2: To write the cylindrical region $E^{*}$.

It can be written in cylindrical coordinates as $$E^{*}=\left\{(r,\theta,z): (r,\theta)\in D,\, r^{2}\leqslant z\leqslant \sqrt{r^{2}}\right\},$$ with the polar region $D$.

Step 3: To know the polar region $D$.

To know $D$, we have to work with $(*)$, we can help ourselves with a plot as usual.

From it, we can see that $\theta$ runs since $-\pi/2$ (or $3\pi/2)$ to $\pi/2$ and the radius $r$ runs since $0$ to $1$. Thus, $$D=\left\{(r,\theta): 0\leqslant r\leqslant 1, 3\pi/2\leqslant \theta\leqslant \pi/2\right\}$$ or equivalentely $$D=\left\{(r,\theta): 0\leqslant r\leqslant 1, -\pi/2\leqslant \theta\leqslant \pi/2\right\}$$

Step 4: To apply the change of variables theorem in terms of cylindrical coordinates.

The change of variables theorem says that $$\iiint_{E}f(x,y,z)\, dV=\iiint_{E^{*}}f(r,\theta,z)\, |J(r,\theta,z)|\,dzdrd\theta,$$ where $J(r,\theta,z)$ is the Jacobian of the change of variables, that in this case we have $J(r,\theta,z)=r$.

Step 5: Find the multiple integral of $f$ over $E\to E^{*}$.

In this case,

\begin{align*}\int_{-1}^{1}\int_{0}^{\sqrt{1-y^{2}}}\int_{x^{2}+y^{2}}^{\sqrt{x^{2}+y^{2}}}xyz\, dzdxdy&=\int_{3\pi/2}^{\pi/2}\int_{0}^{1}\int_{r^{2}}^{r}r^{3}\cos\theta\sin\theta z\, dzdrd\theta\\ &=\int_{-\pi/2}^{\pi/2}\int_{0}^{1}\int_{r^{2}}^{r}r^{3}\cos\theta\sin\theta z\, dzdrd\theta\\ &=0 \end{align*}

NB: Of course, we could have thought of an argument by symmetry to conclude that the answer is $0$, which in this case is feasible.

Answer 2

I use geogebra to visualize this usually. Here are some pictures

From birds eye view of the zed axis, we have

From the side, we have

As we can see, our region of integration is simply the volume between the cone and paraboloid, cut in half along the y axis (green).

To convert this to cylindrical coordinates, we will set the bounds from zed, then to $r$, then to theta.

For zed, we simply will be using the paraboloid and cone as our bounds. We can see this as follows. Draw a line extending from $z=-\infty$ to $\infty$ within the half circle that we have seen previously, and see what it passes through. The line will pass through the paraboloid first, then the cone

However, we have to swap this for our new variables though (ie from x, y to r, theta), so we simply have to use $\sqrt{x^2+y^2}=r$ per the definition to turn our lower bound and upper bound for $z$ to $r^2$ and $r$, respectively.

Now, we can completely get rid of the zed axis and look at the $x-y$ plane in polar coordinates. Our region of integration is as follows (in yellow), which is the 2D projection of the intersection between the paraboloid and the cone.

To define the r bounds, notice that we start at the origin, and no matter what arrow we draw from the origin to the outer bound, the magnitude of the arrow is always one. Hence, the r bounds are $0$ and $1$.

Now to define theta, we can use our $x$ axis as $0$ radians, and go clockwise and counterclockwise from there to find the bounds, which are $-\pi\over2$ and $\pi\over2$

Of course, since there are multiple ways to represent the same area in polar coordinates, you can always use those as well and they will give the same answer.

Of course, for our cylindrical coordinate transformation $f(z, y, x)\mapsto f(r\cos\theta, r\sin\theta, z)$, our jacobian can be trivially calculated to be $r$.

Hence, the integral you give is simply $$\int_{-1}^1 \int_{0}^{\sqrt{1-y^2}}\int_{x^2+y^2}^{\sqrt{x^2+y^2}} xyz dz dx dy = \boxed{\int_{-{\pi\over2}}^{\pi\over2} \int_0^1 \int_{r^2}^r r\cos\theta\cdot r\sin\theta\cdot z\cdot r dzdrd\theta}$$

If we were to evaluate this, we would get $0$.